Question

A point moves in such a way that the sum of squares of its distance from the points $A(2,0)$ and $B( - 2,0)$ is always equal to the square of the distance between A and B. Then find the locus of the point.
A.${x^2} + {y^2} - 2 = 0$
B. ${x^2} + {y^2} + 2 = 0$
C. ${x^2} + {y^2} + 4 = 0$
D. ${x^2} + {y^2} - 4 = 0$

Answer 1

Hint: First suppose the coordinate of the point. Then use the distance formula to obtain the distance between the point and A. After that, obtain the distance between the point and B. Then obtain the distance from A to B. Substitute the obtained values in the given condition and simplify to obtain the answer.

Formula Used:
The distance between the points $(a,b)$ and $(c,d)$ is
$\sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $ .
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${(a - b)^2} = {a^2} - 2ab + {b^2}$

Complete step by step solution:
Suppose the point is $P(x,y)$ .
The given points are$A(2,0)$ and $B( - 2,0)$.
The mathematical equation of the given condition is,
 $P{A^2} + P{B^2} = A{B^2}$ ---(1)
The distance between P and A is,
$PA = \sqrt {{{\left( {2 - x} \right)}^2} + {y^2}} $
$\therefore P{A^2} = {\left( {2 - x} \right)^2} + {y^2}$
The distance between P and B is,
$PB = \sqrt {{{\left( { - 2 - x} \right)}^2} + {y^2}} $
$\therefore P{B^2} = {\left( {2 + x} \right)^2} + {y^2}$
The distance between A and B is,
$AB = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {0^2}} $
$\therefore A{B^2} = {4^2} = 16$
Substitute the obtained values of $P{A^2},P{B^2},A{B^2}$ in the equation (1) and simplify to obtain the required result.
${\left( {2 - x} \right)^2} + {y^2} + {(2 + x)^2} + {y^2} = 16$
$4 - 4x + {x^2} + {y^2} + 4 + 4x + {x^2} + {y^2} = 16$
$2({x^2} + {y^2}) = 8$
${x^2} + {y^2} = 4$

Option ‘D’ is correct

Note: Write the mathematical equation stated in the question for clarification and then solve the problem. The above result shows that the locus of the point whose distance from a fixed line is square the distance from a fixed point is a circle.